dc choppers

1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

DC Choppers


Introduction• Chopper is a static device.

• A variable dc voltage is obtained from a constant dc voltage source.

• Also known as dc-to-dc converter.

• Widely used for motor control.

• Also used in regenerative braking.

• Thyristor converter offers greater efficiency, faster response, lower maintenance, smaller size and smooth control.


Choppers are of Two Types

Step-down choppers. Step-up choppers.

In step down chopper output voltage is less than input voltage.

In step up chopper output voltage is more than input voltage.


Principle Of Step-down Chopper

V

i0

V 0

C hopper

R

+


• A step-down chopper with resistive load.

• The thyristor in the circuit acts as a switch.

• When thyristor is ON, supply voltage appears across the load

• When thyristor is OFF, the voltage across the load will be zero.


V d c

v 0

V

V /R

i0

Id c

t

t

tO N

T

tO F F


verage value of output or load voltage.

verage value of output or load current.

Time interval for which SCR conducts.

Time interval for which SCR is OFF.

Period of switching

dc

dc

ON

OFF

ON OFF

V A

I A

t

t

T t t

or chopping period.

1 Freq. of chopper switching or chopping freq.f

T


Average Output Voltage

.

duty cycle

ONdc

ON OFF

ONdc

ON

tV V

t t

tV V V d

T

tbut d

t


2

0

Average Output Current

RMS value of output voltage

1 ON

dcdc

ONdc

t

O o

VI

RtV V

I dR T R

V v dtT


2

0

2

But during ,

Therefore RMS output voltage

1

.

.

ON

ON o

t

O

ONO ON

O

t v V

V V dtT

tVV t V

T T

V d V


2

2

Output power

But

Output power

O O O

OO

OO

O

P V I

VI

R

VP

R

dVP

R


Effective input resistance of chopper

The output voltage can be varied by

varying the duty cycle.

idc

i

VR

I

RR

d

3.1 Basic DC to DC converters 3.1.1Buck converterSPDT switch changes dc

component

Switch output voltage waveform

Duty cycle D: 0 ≤ D ≤ 1

complement D: D´ = 1 - D

+ -

+

-

V(t)

R

Vg

+

-

Vs(t)

Vs(t) Vg

switch position:

DTs

D Ts

0

1

1

2

t

1

2

0

DTs

Ts

Dc component of switch output voltage

Vs(t) Vg

<Vs>=DVg

DTs

Ts

0

t

Fourier analysis:DC component =average value:

0

area= D Ts Vg

<Vs>

=

∫

0

Ts Vs(t)

dt

Ts

1

<Vs>

=

=DVg

1

Ts DTsVg

Insertion of low- pass filter to remove switching harmonics and pass only dc component

+ -

+

-

V(t)

R

Vg

+

-

Vs(t)

1

2

L

C

v≈<vs>

=DVg

V

Vg

o

0 1 D

• Basic operation principle of buck converter

+ -

+

-

V(t)

R

Vg

+

-

Vs(t)

1

2

L

C

Buck converter with

ideal switch

Realization using

power MOSFET

and diode

+ -

+

-

VL(t)

ic(t)

Vg

iL(t)

t

DTs

Ts

+

L

D1

R

Thought process in analyzing basic DC/DC converters

1) Basic operation principle (qualitative analysis)

–How does current flows during different switching states

–How is energy transferred during different switching states

2) Verification of small ripple approximation

3) Derivation of inductor voltage waveform during different switching states

4) Quantitative analysis according to inductor volt-second balance or capacitor charge balance

• Actual output voltage waveform of buck converter

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

1

2

L

C

Buck converter

containing practical

low-pass filter

ic(t)

iL(t)

Actual output voltage

waveform

v(t ) = V + v ripple(t)

v(t )

V

0

t

Actual waveform

v(t ) = V + v ripple(t)

DC component V

• Buck converter analysis: inductor current waveform

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

1

2

L

C

original

converter

ic(t)

iL(t)

Switch in position 1


+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

• Inductor voltage and current subinterval 1: switch in position 1

Inductor voltage

vL=Vg - v(t) Small ripple approximation:

vL=Vg - V

Knowing the inductor voltage, we can now find the inductor current via

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

vL(t)=L diL(t)

dt Solve for the slope:

diL(t)

dt =

vL(t)

L ≈ Vg - V

L

the inductor current changes with an

essentially constant slope

• Inductor voltage and current subinterval 2: switch in position 2

Inductor voltage

vL= - v(t) Small ripple approximation:

vL ≈ - V

Knowing the inductor voltage, we can now find the inductor current via

vL(t)=L diL (t)

dt Solve for the slope:

diL(t)

dt ≈

V L

the inductor current changes with an

essentially constant slope -

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

Inductor voltage and current waveforms

VL(t)

Vg -V

switch position:

DTs

DTs

-V

1

1

2

t

vL(t)=L diL (t)

dt iL(t)

t

DTs

Ts

0

I

iL(0)

iL(DTs)

Vg -V

L

-V

L

Δ iL

Determination of inductor current ripple magnitude

（changes in iL）=（slope）（length of subinterval）

Vg -V

L

2Δ iL

DTs

=

Δ iL

=

Vg -V

2L DTs

L

=

Vg -V

2Δ iL

DTs

iL(t)

DTs

Ts

0

I

iL(0)

iL(DTs)

Vg -V

L

-V

L

Δ iL

Inductor current waveform during start-up transient

iL(t)

t

DTs

Ts

0

When the converter operates in equilibrium:

iL(0)=0

iL(Ts)

Vg –v(t)

L

-v(t)

L

2Ts

iL(nTs)

nTs

(n+1)Ts

iL((n+1)Ts)

iL((n+1)Ts)= iL(nTs)

• The principle of inductor volt- second balance: DerivationInductor defining relation:

Integrate over one complete switching period:

In periodic steady state, the net changes in inductor current is zero:

Hence, the total area(or volt-seconds)under the inductor voltage waveform is zero whenever the converter operates in steady state.

An equivalent form:

The average inductor voltage is zero in steady state.

vL(t)=L diL (t)

dt

∫ 0

Ts VL(t)

dt

L 1

iL(Ts) -

iL(0)=

∫ 0

Ts VL(t)

dt

=

0

<vL>

=

∫ 0

Ts VL(t)

dt

Ts 1

=

0

• Inductor volt-second balance:Buck converter example

Integral of voltage waveform is area of rectangles:

average voltage is

Equate to zero and solve for V:

inductor voltage waveform

previously derived:

VL(t)

Vg -V

DTs

-V

t

total area λ

∫ 0

Ts VL(t)

dt

=

λ

= (Vg –V)( DTs)+( -V) ( DTs)

<vL>

=

Ts λ

=D (Vg –V) +D'( -V)

0=D Vg –(D+D')V= D Vg –V

V=D Vg

3.1.2Boost converter

• Boost converter example

+ -

+

-

v

R

Vg

+

-

vL(t)

1

2

L

C

Boost converter

with ideal switch

iL(t)

iC(t)

Realization using

power MOSFET

and diode

+ -

ic(t)

Vg

iL(t)

t

DTs

Ts

+

-

VL(t)

L

D1

R

+ -

Q1

+

-

v

C

• Boost converter analysis

original

converter



+ -

+

-

v

R

Vg

+

-

vL(t)

1

2

L

C

iL(t)

+ -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

iC(t)

+ -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

• Subinterval 1: switch in position 1

Inductor voltage and capacitor current

vL=Vg

Small ripple approximation:

iC= - v/R + -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

vL=Vg

iC= - V/R

• Subinterval 2: switch in position 2

Inductor voltage and capacitor current

vL=Vg -v

Small ripple approximation:

iC=iL - v/R

vL=Vg -V

iC= I - V/R

+ -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

• Inductor voltage and capacitor current waveforms

VL(t)

Vg

DTs

D'Ts

Vg -V

t

iC(t)

-V/R

DTs

D'Ts

1 –V/R

t

• Inductor volt- second balance

VL(t)

Vg

DTs

D'Ts

Vg -V

t

∫

0

Ts VL(t)

dt

= ( Vg) DTs+(Vg –V) D'Ts

Net volt-seconds applied to inductor

over one switching period：

Equate to zero and collect terms：

Vg（D+ D'）-VD'=0

Solve for V：

V=

Vg

D'

The voltage conversion ratio is therefore：

V

Vg

D'

M（D）=

=

=

1

1-D

1

Conversion ratio M(D) of the boost converter

D'

M（D）=

=

1

1-D

1

D

M（D）

0

0

1

2

3

4

5

0.2

0.4

0.6

0.8

1

• Determination of inductor current dc component

Vg/R

I

D

0

0

2

4

6

8

0.2

0.4

0.6

0.8

1

iC(t)

-V/R

DTs

D'Ts

I –V/R

t

Capacitor charge balance：

∫

0

Ts iC(t)

dt

=（-

）D'Ts

V

R

）DTs

+（I-

V

R

Collect terms and equate to zero：

-

V

R

（D+D'）+I D'=0

Solve for I：

V

D'R

I=

Eliminate V to express in terms of Vg：

Vg

D' I=

2

R

Continuous- Conduction- Mode (CCM) and Discontinuous Conduction-Mode (DCM) of boost

M E

VD L

V uo EM

a)


Methods Of Control

• The output dc voltage can be varied by the following methods.

– Pulse width modulation control or constant frequency operation.

– Variable frequency control.


Pulse Width Modulation

• tON is varied keeping chopping frequency ‘f’ & chopping period ‘T’ constant.

• Output voltage is varied by varying the ON time tON


V 0

V

V

V 0

t

ttO N

tO N tO F F

tO F F

T


Variable Frequency Control

• Chopping frequency ‘f’ is varied keeping either tON or tOFF constant.

• To obtain full output voltage range, frequency has to be varied over a wide range.

• This method produces harmonics in the output and for large tOFF load current may become discontinuous


v 0

V

V

v 0

t

t

tO N

tO N

T

T

tO F F

tO F F


Step-down ChopperWith R-L Load

V

i0

V 0

C hopper

R

LFW D

E

+


• When chopper is ON, supply is connected across load.

• Current flows from supply to load.

• When chopper is OFF, load current continues to flow in the same direction through FWD due to energy stored in inductor ‘L’.


• Load current can be continuous or discontinuous depending on the values of ‘L’ and duty cycle ‘d’

• For a continuous current operation, load current varies between two limits Imax and Imin

• When current becomes equal to Imax the chopper is turned-off and it is turned-on when current reduces to Imin.


O utpu tvo ltage

O utpu tcurrent

v 0

V

i0

Im a x

Im in

t

t

tO N

T

tO F F

C ontinuouscurrent

O utpu tcurrent

t

D iscon tinuouscurrent

i0


Expressions For Load Current

iO For Continuous Current Operation When

Chopper Is ON (0 t tON)


V

i0

V 0

R

L

E

+

-


min

min

Taking Laplace Transform

. 0

At 0, initial current 0

OO

O O O

O

O

diV i R L E

dt

V ERI S L S I S i

S S

t i I

IV EI S

RR SLS SLL


min

Taking Inverse Laplace Transform

1

This expression is valid for 0 ,

i.e., during the period chopper is ON.

At the instant the chopper is turned off,

load c

R Rt t

L LO

ON

V Ei t e I e

R

t t

maxurrent is O ONi t I


When Chopper is OFF

i0

R

L

E


max

When Chopper is OFF 0

0

Talking Laplace transform

0 0

Redefining time origin we have at 0,

initial current 0

OFF

OO

O O O

O

t t

diRi L E

dt

ERI S L SI S i

St

i I


max

max

Taking Inverse Laplace Transform

1

O

R Rt t

L LO

I EI S

R RS LS SL L

Ei t I e e

R


min

The expression is valid for 0 ,

i.e., during the period chopper is OFF

At the instant the chopper is turned ON or at

the end of the off period, the load current is

OFF

O OFF

t t

i t I


min

max

max

max min

min

From equation

1

At ,

To Find &

1

R Rt t

L LO

ON O

dRT dRT

L L

V Ei t e I e

R

t t dT i t I

V EI e I e

I I

R


max

min

From equation

1

At ,

1

R Rt t

L LO

OFF ON O

OFF

Ei t I e e

R

t t T t i t I

t t d T


1 1

min max

min

max min

max

1

Substituting for in equation

1

we get,

1

1

d RT d RT

L L

dRT dRT

L L

dRT

L

RT

L

EI I e e

R

I

V EI e I e

R

V e EI

R Re


max

1 1

min max

min

max min

Substituting for in equation

1

we get,

1

1

is known as the steady state ripple.

d RT d RT

L L

dRT

L

RT

L

I

EI I e e

R

V e EI

R Re

I I


max min

max min

Therefore peak-to-peak ripple current

Average output voltage

.

Average output current

2

dc

dc approx

I I I

V d V

I II


min max

min

max minmin

Assuming load current varies linearly

from to instantaneous

load current is given by

. 0O ON

O

I I

I ti I for t t dT

dTI I

i I tdT


20

0

2

max minmin

0

2min max min2 2max min

min

0

RMS value of load current

1

1

21

dT

O RMS

dT

O RMS

dT

O RMS

I i dtdT

I I tI I dt

dT dT

I I I tI II I t dt

dT dT dT


12 2

max min2min min max min

20

0

2

max minmin

0

RMS value of output current

3

RMS chopper current

1

1

O RMS

dT

CH

dT

CH

I II I I I I

I i dtT

I II I t dt

T dT


12 2

max min2min min max min3

Effective input resistance is

CH

CH O RMS

iS

I II d I I I I

I d I

VR

I


Where

Average source currentS

S dc

idc

I

I dI

VR

dI


Principle Of Step-up Chopper

+

V OV

C hopper

CLOAD

DLI

+


• Step-up chopper is used to obtain a load voltage higher than the input voltage V.

• The values of L and C are chosen depending upon the requirement of output voltage and current.

• When the chopper is ON, the inductor L is connected across the supply.

• The inductor current ‘I’ rises and the inductor stores energy during the ON time of the chopper, tON.


• When the chopper is off, the inductor current I is forced to flow through the diode D and load for a period, tOFF.

• The current tends to decrease resulting in reversing the polarity of induced EMF in L.

• Therefore voltage across load is given by

. ., O O

dIV V L i e V V

dt


• A large capacitor ‘C’ connected across the load, will provide a continuous output voltage .

• Diode D prevents any current flow from capacitor to the source.

• Step up choppers are used for regenerative braking of dc motors.


Expression For Output Voltage

Assume the average inductor current to be

during ON and OFF time of Chopper.

Voltage across inductor

Therefore energy stored in inductor

= . .

Where

When Chopper

period of chopper.

is ON

ON

ON

I

L V

V I t

t ON


(energy is supplied by inductor to load)

Voltage across

Energy supplied by inductor

where period of Chopper.

Neg

When Chopper

lecting losses, energy stored in inductor

is OFF

O

O OFF

OFF

L V V

L V V It

t OFF

L

= energy supplied by inductor L


Where

T = Chopping period or period

of switching.

ON O OFF

ON OFFO

OFF

OON

VIt V V It

V t tV

t

TV V

T t


1

1

1

1

Where duty cyle

ON OFF

OON

O

ON

T t t

V Vt

T

V Vd

td

T


For variation of duty cycle ' ' in the

range of 0 1 the output voltage

will vary in the range O

O

d

d V

V V


Performance Parameters• The thyristor requires a certain minimum time to

turn ON and turn OFF.

• Duty cycle d can be varied only between a min. & max. value, limiting the min. and max. value of the output voltage.

• Ripple in the load current depends inversely on the chopping frequency, f.

• To reduce the load ripple current, frequency should be as high as possible.


Classification Of Choppers

• Choppers are classified as – Class A Chopper– Class B Chopper– Class C Chopper– Class D Chopper– Class E Chopper


Class A Chopper

V

C hopper

FW D

+

v 0

v 0

i0

i0

LOAD

V


• When chopper is ON, supply voltage V is connected across the load.

• When chopper is OFF, vO = 0 and the load current continues to flow in the same direction through the FWD.

• The average values of output voltage and current are always positive.

• Class A Chopper is a first quadrant chopper .


• Class A Chopper is a step-down chopper in which power always flows form source to load.

• It is used to control the speed of dc motor.

• The output current equations obtained in step down chopper with R-L load can be used to study the performance of Class A Chopper.


O utpu t cu rren t

Thyristo rgate pu lse

O utpu t vo ltage

ig

i0

v 0

t

t

ttO N

T

C H O N

FW D C onduc ts


Class B Chopper

V

C hopper

+

v 0

v 0

i0

i0

L

E

R

D


• When chopper is ON, E drives a current through L and R in a direction opposite to that shown in figure.

• During the ON period of the chopper, the inductance L stores energy.

• When Chopper is OFF, diode D conducts, and part of the energy stored in inductor L is returned to the supply.


• Average output voltage is positive.• Average output current is negative. • Therefore Class B Chopper operates in second

quadrant.• In this chopper, power flows from load to

source.• Class B Chopper is used for regenerative

braking of dc motor.• Class B Chopper is a step-up chopper.


O utpu t cu rren t

D con d u cts C h o pp e r

con d u cts

Thyristo rgate pu lse

O utpu t vo ltage

ig

i0

v 0

t

t

t

Im in

Im ax

T

tO NtO F F


Expression for Output Current


min

For the initial condition i.e.,

During the interval diode 'D' conduc

at 0

The solution of the ab

ts

voltage equation

ove equation is obtained

along similar lines as in s

is given by

OO

O

LdiV Ri E

dt

i t I t

tep-down chopper

with R-L load


min

max

max min

During the interval chopper is ON voltage

equation is g

1 0

At

1

0

iven by

OFF OFF

R Rt t

L LO OFF

OFF O

R Rt t

L L

OO

V Ei t e I e t t

R

t t i t I

V EI e I e

R

LdiRi E

dt


max

max

min

min max

Redefining the time origin, at 0

The solution for the stated initial condition is

1 0

At

1ON ON

O

R Rt t

L LO ON

ON O

R Rt t

L L

t i t I

Ei t I e e t t

R

t t i t I

EI I e e

R


Class C Chopper

V

C hopper

+

v 0

D 1

D 2C H 2

C H 1

v 0i0

i0

L

E

R


• Class C Chopper is a combination of Class A and Class B Choppers.

• For first quadrant operation, CH1 is ON or D2 conducts.

• For second quadrant operation, CH2 is ON or D1 conducts.

• When CH1 is ON, the load current is positive.• The output voltage is equal to ‘V’ & the load

receives power from the source. • When CH1 is turned OFF, energy stored in

inductance L forces current to flow through the diode D2 and the output voltage is zero.


• Current continues to flow in positive direction.• When CH2 is triggered, the voltage E forces

current to flow in opposite direction through L and CH2 .

• The output voltage is zero.• On turning OFF CH2 , the energy stored in the

inductance drives current through diode D1 and the supply

• Output voltage is V, the input current becomes negative and power flows from load to source.


• Average output voltage is positive• Average output current can take both positive

and negative values.• Choppers CH1 & CH2 should not be turned

ON simultaneously as it would result in short circuiting the supply.

• Class C Chopper can be used both for dc motor control and regenerative braking of dc motor.

• Class C Chopper can be used as a step-up or step-down chopper.


G ate pu lseof C H 2


O utput cu rren t

O utpu t vo ltage

ig 1

ig 2

i0

V 0

t

t

t

t

D 1 D 1D 2 D 2C H 1 C H 2 C H 1 C H 2

O N O N O N O N


Class D Chopper

V+ v 0

D 2

D 1 C H 2

C H 1

v 0

i0

L ER i0


• Class D is a two quadrant chopper.

• When both CH1 and CH2 are triggered simultaneously, the output voltage vO = V and output current flows through the load.

• When CH1 and CH2 are turned OFF, the load current continues to flow in the same direction through load, D1 and D2 , due to the energy stored in the inductor L.

• Output voltage vO = - V .


• Average load voltage is positive if chopper ON time is more than the OFF time

• Average output voltage becomes negative if tON < tOFF .

• Hence the direction of load current is always positive but load voltage can be positive or negative.




O utpu t curren t

O utpu t vo ltage

Average v 0

ig 1

ig 2

i0

v 0

V

t

t

t

t

C H ,C HO N1 2 D 1,D 2 Conducting




O utpu t curren t

O utpu t vo ltage

Average v 0

ig 1

ig 2

i0

v 0

V

t

t

t

t

C HC H

1

2

D , D1 2


Class E Chopper

V

v 0

i0L ER

C H 2 C H 4D 2 D 4

D 1 D 3C H 1 C H 3

+


Four Quadrant Operationv 0

i0

C H - C H O NC H - D C onduc ts

1 4

4 2

D D2 3 - C onductsC H - D C onduc ts4 2

C H - C H O NC H - D C onduc ts

3 2

2 4

C H - D C onduc tsD - D C onducts

2 4

1 4


• Class E is a four quadrant chopper

• When CH1 and CH4 are triggered, output current iO flows in positive direction through CH1 and CH4, and with output voltage vO = V.

• This gives the first quadrant operation.

• When both CH1 and CH4 are OFF, the energy stored in the inductor L drives iO through D2 and D3 in the same direction, but output voltage vO = -V.


• Therefore the chopper operates in the fourth quadrant.

• When CH2 and CH3 are triggered, the load current iO flows in opposite direction & output voltage vO = -V.

• Since both iO and vO are negative, the chopper operates in third quadrant.


• When both CH2 and CH3 are OFF, the load current iO continues to flow in the same direction D1 and D4 and the output voltage vO = V.

• Therefore the chopper operates in second quadrant as vO is positive but iO is negative.


Impulse Commutated Chopper

• Impulse commutated choppers are widely used in high power circuits where load fluctuation is not large.

• This chopper is also known as

– Parallel capacitor turn-off chopper

– Voltage commutated chopper

– Classical chopper.


LOAD

L

C

IL

L S

V S

+

_

+

_

T 2

T 1

D 1

a

biC

iT 1

v O

+

_

FW D


• To start the circuit, capacitor ‘C’ is initially charged with polarity (with plate ‘a’ positive) by triggering the thyristor T2.

• Capacitor ‘C’ gets charged through VS, C, T2 and load.

• As the charging current decays to zero thyristor T2 will be turned-off.

• With capacitor charged with plate ‘a’ positive the circuit is ready for operation.

• Assume that the load current remains constant during the commutation process.


• For convenience the chopper operation is divided into five modes.– Mode-1

– Mode-2

– Mode-3

– Mode-4

– Mode-5


Mode-1 Operation

LOADL

C

IL

L S

V S

+

_

+

_

T 1

D 1

V C iC


• Thyristor T1 is fired at t = 0.

• The supply voltage comes across the load.

• Load current IL flows through T1 and load.

• At the same time capacitor discharges through T1, D1, L1, & ‘C’ and the capacitor reverses its voltage.

• This reverse voltage on capacitor is held constant by diode D1.


Capacitor Discharge Current

sin

1Where

& Capacitor Voltage

cos

C

C

Ci t V t

L

LC

V t V t


Mode-2 Operation

LOAD

C

L S

V S+

_+

_

T 2

V C

IL

IL


• Thyristor T2 is now fired to commutate thyristor T1.

• When T2 is ON capacitor voltage reverse biases T1 and turns if off.

• The capacitor discharges through the load from –V to 0.

• Discharge time is known as circuit turn-off time.


C

Circuit turn-off time is given by

Where is load current.

t depends on load current, it must be designed

for the worst case condition which occur at the

maximum value of load current and mini

CC

L

L

V Ct

I

I

mum

value of capacitor voltage.


• Capacitor recharges back to the supply voltage (with plate ‘a’ positive).

• This time is called the recharging time and is given by

• The total time required for the capacitor to discharge and recharge is called the commutation time and it is given by

Sd

L

V Ct

I

r C dt t t


• At the end of Mode-2 capacitor has recharged to VS and the free wheeling diode starts conducting.


Mode-3 Operation

LOAD

C

L S

V S

+_

+

_

T 2V S

FW D

IL

IL


• FWD starts conducting and the load current decays.

• The energy stored in source inductance LS is transferred to capacitor.

• Hence capacitor charges to a voltage higher than supply voltage, T2 naturally turns off.


The instantaneous capacitor voltage is

sin

Where

1

SC S L S

S

S

LV t V I t

C

L C


Mode-4 Operation

LOAD

C

L S

V S

+_

+

_

D 1

LFW D

IL

V C


• Capacitor has been overcharged i.e. its voltage is above supply voltage.

• Capacitor starts discharging in reverse direction.

• Hence capacitor current becomes negative. • The capacitor discharges through LS, VS, FWD,

D1 and L. • When this current reduces to zero D1 will stop

conducting and the capacitor voltage will be same as the supply voltage


Mode-5 Operation

• Both thyristors are off and the load current flows through the FWD.

• This mode will end once thyristor T1 is fired.

LOAD

IL

FW D


C a p a c i to r C u rr e n tI L

t

t

I p C u r r e n t t h r o u g h T 1

ic

0I p

iT1

0

I L


t

t

t

Vo l ta g e a c r o s s T 1

O u tp u t Vo l t a g e

C a p a c i to r Vo l t a g e

tctd

v T1

V c

0v o

V s c+ V

V s

v c

V c

- V c


Disadvantages• A starting circuit is required and the starting circuit

should be such that it triggers thyristor T2 first.

• Load voltage jumps to almost twice the supply voltage when the commutation is initiated.

• The discharging and charging time of commutation capacitor are dependent on the load current and this limits high frequency operation, especially at low load current.


• Chopper cannot be tested without connecting load.

• Thyristor T1 has to carry load current as well as resonant current resulting in increasing its peak current rating.

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dc choppers